Problem of the Week

The COMC has three parts. In part A solutions do not require work be shown and may be possible to do in your head. In part B the problems begin to draw on more knowledge and have some more challenging aspects that will need a pencil and paper to solve. By Part C the problems require that work be shown and involve arguments to support the answer.

We have selected a cross section of contest problems from a variety of national and regional contests that we hope will stimulate interest in problem solving and give some experience to get ready to write the Canadian Open Mathematics Competition in November. The problem areas are not tied to particular grade levels, or to the curriculum but cover a number of areas from algebra, through logic and some geometry.

We will post solutions to these problems one week later, but teachers should be aware that determined students may be able to locate solutions elsewhere online before then.

For a more comprehensive set of problems and solutions at each of these levels, please feel welcome to download past official exams and solutions from our archive.

Need more COMC Problems-of-the-Week? Take a look at the set from past years:
2016,
2015,
2014,
or 2013!

This is problem 4 of the Georgian Math Olympiad 1997, Grade X, which appeared in Crux Mathematicorum in [2000:193]. We give the solution of Amengual Covas that appeared in Crux Mathematicorum in [2002:204-205], modified by the editor.

An acute triangle $ABC$ is inscribed in a circle with the center at $O$. The bisector of $\angle A$ meets the side $BC$ at $D$. The perpendicular from $D$ to $AO$ meets the side $AC$ at a point $P$ which is interior to $AC$. Show that $AB=AP$.

This is problem 5 of the 1995 Italian Math Olympiad which appeared in Crux Mathematicorum in [1998:323-324]. We give the solution of Pierre Bornsztein that appeared in Crux Mathematicorum in [2000:79-80].

Since $ABC$ is acute, $O$ is interior to $ABC$. Since $P$ is interior to $AC$, $O$ is interior to $ADC$.

This is problem 4 of the Ukrainian Math Olympiad, 11th Grade, which appeared in Crux Mathematicorum in [2006:217-218]. We give the solution of Michel Bataille that appeared in Crux Mathematicorum in [2007:227-228].

There is no such pair $(p,q)$. To prove this, we argue by contradiction.

Let $P(X)$ and $Q(X)$ be two polynomials with integer coefficients. If $\frac{P(n)}{Q(n)}$ is an integer for every integer $n$, prove that there exists a polynomial $S(X)$ with rational coefficients such that $P(X)=Q(X)S(X)$.

This is problem 5 of Alberta High School Math Competition 1995, Part B. We present the official solution.

We divide $P(X)$ by $Q(X)$ and obtain a quotient $S(X)$ and a remainder $R(X)$. Then, we have
$$P(X)=Q(X)S(X)+R(X) \,,$$
with $\deg(R(X)) < \deg(Q(X))$. We then have:
$$
\begin{align}
\frac{P(X)}{Q(X)}&= S(X)+ \frac{R(X)}{Q(X)} \,.\tag{1}\cr
\end{align}
$$

Since both $P(X)$ and $Q(X)$ have integer coefficients, $S(X)$ has rational coefficients. To complete the proof we show that $R(X)=0$.

We can write $$S(X)=\frac{1}{m} T(X) \,,$$ for some integer $m > 0$ and some polynomial with integer coefficients. By combining the hypothesis with (1) we then get that for all integers $n$ we have
$$m \frac{R(n)}{Q(n)} = m\frac{P(n)}{Q(n)}-T(n) \in \mathbb Z \,.$$

Since $\deg(Q(x)) > \deg(R(x))$, for $x$ sufficiently large we have
$$\left| m \frac{R(x)}{Q(x)} \right| < 1\,.$$

Therefore, for all integers $n$ which are sufficiently large we get that $\left| m \frac{R(n)}{Q(n)} \right|=0$ and hence, that $R(n)=0$. This shows that $R$ has infinitely many roots, and thus $R=0$, as claimed.

Editor note: Note that we cannot conclude that $S(x)$ has integer coefficients. Indeed, if $P(X)=X^2+X$ and $Q(X)=2$ then $\frac{P(n)}{Q(n)}=\frac{n(n+1)}{2}$ is an integer for all integers $n$ but $S(X)=\frac{1}{2}X^2+\frac{1}{2}$.

Pólya proved that a polynomial $P(X)$ takes integer values at all integers if and only if there exists some $n$ and integers $a_0,..,a_n$ such that
$$
\begin{align}
P(X)&= a_n \frac{X(X-1)...(X-n+1)}{n!}+a_{n-1} \frac{X(X-1)+...+(X-n+2)}{(n-1)!}+...\cr
&+a_{2} \frac{X(X-1)}{2!}+a_1X+a_0 \,.\cr
\end{align}
$$

This is problem 5 of the 1999 Íslenzka Staerŏfræŏikeppni Framhaldsskólanema which appeared in Crux Mathematicorum in [2001:232-233]. We give the solution of Michel Bataille that appeared in Crux Mathematicorum in [2003:304].

It is easy to see that $a_n >0$ for all $n$. Then, for all $n$ we have $$
\frac{1}{a_{n+1}}=\frac{1}{a_n}+n \,.$$

Problem 5 of the $21^{\mbox{st}}$ Austian Mathematical Olympiad, Final Round, 1990, which appeared in Crux Mathematicorum [1992:100]. We present the similar solutions by Joseph Ling, Pavlos Maragoudakis and Michael Selby which appeared at [1993:138].

If $r=0$ then the equation becomes $x-1=0$, so $x=1$. Therefore $r=0$ is a solution.

Problem 10 of the British Columbia Secondary School Mathematics Contest, 2008, Junior Final, Part A which appeared in the Skoliad Corner of Crux Mathematicorum at [2008:321-324]. We present the solution by Jixuan Wang that appeared at [2009:269].